
#1
Aug1508, 08:43 AM

P: 178

Since the Hermite Polynomials are orthogonal, could one state that they span all polynomials [tex]f[/tex] where [tex]f : R \rightarrow R[/tex]? This would be EXTREMELY useful for the harmonic oscillator potential in quantum mechanics...




#2
Aug1508, 01:14 PM

P: 2,163

(1, 0, 0) and (0, 1, 0) They are orthogonal to each other, but they don't span. on the other hand, consider the three vectors: (1, 0, 0), (1, 1, 0), (1, 1, 1) They span, but they are not orthogonal to each other. Therefore, knowing that the Hermite Polynomials are orthogonal is not enough to show that they span. You would need to prove that. 



#3
Aug1508, 01:28 PM

Sci Advisor
P: 1,185

It doesn't follow from orthogonality alone, but it is nevertheless true.




#4
Aug1508, 02:00 PM

P: 178

Hermite Polynomials 



#5
Aug1508, 02:07 PM

P: 240

They are useful to express any function defined on the range on which they are orthogonal.
Other orthogonal polynomials will be useful to express functions on the range on which they are orthogonal (e.g. Legender polynomials on [1,1]. 



#6
Aug1508, 02:21 PM

Sci Advisor
HW Helper
P: 2,020

In what sense are you using the word "span"? 



#7
Aug1508, 02:27 PM

P: 178

I'm using the word span in the sense that any polynomial of the form
[tex]f(\xi) = a_n \xi^n + a_{n1} \xi^{n1} + a_{n2} \xi^{n2} + \cdots + a_1 \xi + a_0[/tex] can be written as a linear combination of [tex]H_0, H_1, \cdots, H_n[/tex]. I see I had a typo in my last post... I meant that the Hermite polynomials could span all polynomials (real). 



#8
Aug1508, 05:24 PM

P: 2,163

(0,1,0,0,0,...) (0,0,1,0,0,...) (0,0,0,1,0,...) ... 



#9
Aug1508, 05:35 PM

Sci Advisor
HW Helper
Thanks
P: 25,158





#10
Aug1508, 06:00 PM

P: 2,163




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